Answer:
Step-by-step explanation:
To prove that the sum of the first n even +ve integers is:
[tex]\mathsf{2+4+6+8+ . . . +2n = n^2+ n }[/tex]
By using mathematical induction;
For n = 1, we get:
2n = 2 × 1 = 2
2 = 1² + 1 ----- (1)
∴ the outcome is true if n = 1
However, let assume that the result is also true for n = k
Now, [tex]\mathsf{2+4+6+8+. . .+2k = k^2 + k --- (2)}[/tex]
[tex]\mathsf{2+4+6+8+. . .+2k+2(k+1)}[/tex]
we can now say:
[tex]\mathsf{= (k^2 + k) + 2(k + 1)} \\ \\ \mathsf{= k^2 + k + 2k + 1}[/tex]
[tex]\mathsf{= (k^2 + 2k + 1) + (k + 1)}[/tex]
[tex]\mathsf{= (k + 1)^2 + (k + 1)}[/tex]
∴
[tex]\mathsf{2 + 4 + 6 + 8 + . . . + 2k + 2(k + 1) = (k + 1)^2 + (k + 1)}[/tex]
Thus, the result is true for n = m+1, hence we can posit that the result is also true for each value of n.
As such [tex]\mathsf{2+4+6+8+. . .+2n = n^2 + n }[/tex]
